"Determining Gas Volume and Limiting Reagent in a Reaction Between Magnesium and Hydrochloric Acid"

Question

A sample of magnesium ribbon is added to 100.0 mL of 1.00 M hydrochloric acid at 25°C.The reaction is:

ext{MgCl}_2(aq) + ext{H}_2(g)

(a) Write the balanced chemical equation for the reaction.(b) If 0.243 g of magnesium reacts completely, calculate the volume of hydrogen gas produced at STP (0°C and 1 atm).(c) What is the limiting reagent in this reaction?"Determining Gas Volume and Limiting Reagent in a Reaction Between Magnesium and Hydrochloric Acid"

Answer

a) balanced chemical equation the reaction between magnesium metal and hydrochloric acid is:

ext{MgCl}_2(aq) + ext{H}_2(g)

this is already balanced: 1 mg atom on both sides 2 h atoms on both sides 2 cl atoms on both sides answer: mg(s) + 2hcl(aq) → mgcl₂(aq) + h₂(g)b) we are given: mass of mg = 0.243 g molar mass of mg = 24.305 g/mol 1 mol of mg produces 1 mol of h₂ gas at stp, 1 mol of any gas = 22.4 l step 1: moles of mg

\frac{0.243\ ext{g}}{24.305\ ext{g/mol}} = 0.009999\ ext{mol} \approx 0.0100\ ext{mol}

step 2: moles of h₂ produced from the balanced equation: 1 mol mg → 1 mol h₂ so,

h}_2 = 0.0100\ ext{mol}

step 3: volume of h₂ at stp

0.0100\ ext{mol} imes 22.4\ ext{l/mol} = \boxed{0.224\ ext{l}} = \boxed{224\ ext{ml}}

answer : 0.224 l (224 ml) of hydrogen gas produced at stpc) we are given: 100.0 ml of 1.00 m hcl = 0.100 l × 1.00 mol/l = 0.100 mol hcl from earlier: 0.0100 mol mg from the reaction:

hcl1\ ext{mol mg} ext{ reacts with } 2\ ext{mol hcl}

so,

hcl0.0100\ ext{mol mg needs } 0.0200\ ext{mol hcl}

but we have 0.100 mol hcl, which is more than enough.

(mg)\boxed{ ext{limiting reagent: magnesium (mg)}}

answer: limiting reagent: magnesium (mg)

Answer

Answer

solution to magnesium hydrochloric acid reaction problem

given:
100.0 ml of 1.00 m hcl solution
0.243 g mg ribbon
temperature = 25°c (298 k)
stp conditions = 0°c (273 k), 1 atm
molar mass mg = 24.31 g/mol
ideal gas constant r = 0.08206 l·atm/(mol·k)
molar volume at stp = 22.4 l/mol

(a) balanced chemical equation
mg(s) + 2hcl(aq) → mgcl₂(aq) + h₂(g)
(b) volume of h₂ gas produced at stp

step 1: calculate moles of mg
moles mg = mass/molar mass = 0.243 g / 24.31 g/mol = 0.0100 mol
step 2: determine moles h₂ produced (1:1 stoichiometry)
moles h₂ = moles mg = 0.0100 mol
step 3: calculate volume at stp
volume h₂ = moles × molar volume = 0.0100 mol × 22.4 l/mol = 0.224 l

final answer for (b): 0.224 l h₂ gas

(c) limiting reagent analysis
step 1: calculate available moles hcl
moles hcl = m × v = 1.00 mol/l × 0.1000 l = 0.100 mol
step 2: determine required hcl for complete reaction
from stoichiometry: 1 mol mg requires 2 mol hcl
for 0.0100 mol mg: required hcl = 0.0200 mol
step 3: compare available vs. required
available hcl (0.100 mol) > required hcl (0.0200 mol)

conclusion: magnesium is the limiting reagent

verification:
1. mass balance confirms proper stoichiometry
2. volume calculation uses standard stp conditions
3. limiting reagent analysis shows mg is fully consumed first
4. all values reported with proper significant figures

key notes:
- the reaction goes to completion (exothermic)
- stp conditions used for gas volume calculation
- hcl is in excess by 0.080 mol (0.100 - 0.020 mol)

Answer

reaction overview: balanced reaction:

ext{MgCl}_2(aq) + ext{H}_2(g)

🔹 (a) balanced chemical equation

🔹 (b) calculate volume of hydrogen gas produced at stp step 1: moles of magnesium given mass of mg:

molar mass of mg = 24.305 g/mol

step 2: mole ratio from the balanced equation:1 mol mg → 1 mol h₂ so, 0.009998 mol mg → 0.009998 mol h₂ step 3: use ideal gas law at stp at stp (0°c, 1 atm), 1 mol of gas = 22.4 l

🔹 (c) determine the limiting reagent step 1: moles of hcl volume = 100.0 ml = 0.1000 lconcentration = 1.00 m

step 2: required hcl for 0.009998 mol mg mg : hcl = 1 : 2

you have 0.1000 mol hcl, which is more than enough.✅ hcl is in excess🚫 magnesium is the limiting reagent ✅ final answers: (a)

(b)

(c)

Answer

(a) balanced chemical equation
the reaction between magnesium and hydrochloric acid is:
\mathbf{mg(s) + 2hcl(aq) ightarrow mgcl_2(aq) + h_2(g)}
this is already balanced as written.
---
(b) if 0.243 g of magnesium reacts completely, calculate the volume of hydrogen gas produced at stp.
step 1: calculate moles of magnesium.
we need the molar mass of magnesium:
molar mass of mg = 24.305 g/mol
	ext{moles of mg} = \frac{0.243\ 	ext{g}}{24.305\ 	ext{g/mol}} \approx 0.009999\ 	ext{mol}
(we can round this to 0.01000 mol for simplicity.)
step 2: use the stoichiometry of the reaction to find moles of h₂ produced.
from the balanced equation:1 mol mg → 1 mol h₂
so, moles of h₂ produced = moles of mg reacted = 0.01000 mol
step 3: calculate the volume of h₂ gas at stp.
at stp, 1 mole of gas occupies 22.4 l.
v_{h_2} = 0.01000\ 	ext{mol} 	imes 22.4\ 	ext{l/mol} = 0.224\ 	ext{l}
answer:the volume of hydrogen gas produced at stp is 0.224 l (or 224 ml).
---
(c) what is the limiting reagent?
step 1: calculate moles of hcl available.
given:volume of hcl = 100.0 ml = 0.1000 lconcentration of hcl = 1.00 m
	ext{moles of hcl} = 1.00\ 	ext{mol/l} 	imes 0.1000\ 	ext{l} = 0.1000\ 	ext{mol}
step 2: determine how much hcl is needed to react with the given mg.
from the balanced equation:1 mol mg reacts with 2 mol hcl.so, for 0.01000 mol mg:
	ext{required hcl} = 0.01000\ 	ext{mol mg} 	imes 2\ \frac{	ext{mol hcl}}{	ext{mol mg}} = 0.02000\ 	ext{mol hcl}
step 3: compare available vs required hcl.
available hcl: 0.1000 mol
required hcl: 0.02000 mol
since there is more than enough hcl, magnesium is the limiting reagent.
---
final answers:
(a) balanced equation:
\mathbf{mg(s) + 2hcl(aq) ightarrow mgcl_2(aq) + h_2(g)}
(b) volume of hydrogen gas produced: 0.224 l (224 ml) at stp
(c) limiting reagent: magnesium

Answer

answer 
determining gas volume and limiting reagent in a reaction between magnesium and hydrochloric acid
(a) balanced chemical equation:
\text{mg}(s) + 2\text{hcl}(aq) \rightarrow \text{mgcl}_2(aq) + \text{h}_2(g)
(b) if 0.243 g of magnesium reacts completely, calculate the volume of hydrogen gas produced at stp (0°c and 1 atm).
step 1: find the molar mass of mg.
\text{molar mass of mg} = 24.31 \, \text{g/mol}
step 2: calculate moles of mg.
\text{moles of mg} = \frac{0.243 \, \text{g}}{24.31 \, \text{g/mol}} \approx 0.01 \, \text{mol}
step 3: use the mole ratio to find moles of h₂ produced.
from the balanced equation:
1 \text{ mol mg} \rightarrow 1 \text{ mol h}_2
\text{so, 0.01 mol mg} \rightarrow 0.01 \text{ mol h}_2
step 4: calculate volume of h₂ at stp.
at stp, 1 mol of gas = 22.4 l
\text{volume of h}_2 = 0.01 \, \text{mol} \times 22.4 \, \text{l/mol} = \boxed{0.224 \, \text{l}} \, \text{or} \, 224 \, \text{ml}
(c) what is the limiting reagent in this reaction?
we are given:

0.243 g mg → 0.01 mol mg
100.0 ml of 1.00 m hcl →
\text{moles of hcl} = 0.100 \, \text{l} \times 1.00 \, \text{mol/l} = 0.100 \, \text{mol}

from the equation:
$$
1 \text{ mol mg} \reacts \text{ with } 2 \text{ mol hcl}
$$
\text{0.01 mol mg} \times 2 = 0.02 \text{ mol hcl needed}
available hcl = 0.100 mol
required hcl = 0.020 mol
✅ hcl is in excess
⚠️ magnesium is the limiting reagent
final answers summary:
(a) balanced equation:
\text{mg}(s) + 2\text{hcl}(aq) \rightarrow \text{mgcl}_2(aq) + \text{h}_2(g)
(b) volume of hydrogen gas produced at stp:
\boxed{224 \, \text{ml} \, \text{(or 0.224 l)}}
(c) limiting reagent:
\boxed{\text{magnesium (mg)}}

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"Determining Gas Volume and Limiting Reagent in a Reaction Between Magnesium and Hydrochloric Acid"

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